What are the solutions to the equation 2x^3-4x^2-32x+64=0 ?

The solutions to the equation 2x^3-4x^2-32x+64=0 are x=2,x=-4,x=4

Find the zeros of 2x^3-4x^2-32x+64=0

The zeros of 2x^3-4x^2-32x+64=0 are x=2,x=-4,x=4

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2 x^{3} - 4 x^{2} - 32 x + 64 = 0

x = 2, x = - 4, x = 4

Solution steps

2 x^{3} - 4 x^{2} - 32 x + 64 = 0

Factor

2 x^{3} - 4 x^{2} - 32 x + 64 : 2 (x - 2) (x + 4) (x - 4)

2 (x - 2) (x + 4) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 2 = 0 or x + 4 = 0 or x - 4 = 0

Solve

x - 2 = 0 : x = 2

Solve

x + 4 = 0 : x = - 4

Solve

x - 4 = 0 : x = 4

The solutions are

x = 2, x = - 4, x = 4

y^{4} + 2 y^{2} + y = 0

(5 - 2 x) \cdot (x^{2} - 9) = 0

x^{3} = 4.2

2 x^{3} + 4 x^{2} - 30 x = 0

x^{3} = 200

What are the solutions to the equation 2x^3-4x^2-32x+64=0 ?
The solutions to the equation 2x^3-4x^2-32x+64=0 are x=2,x=-4,x=4
Find the zeros of 2x^3-4x^2-32x+64=0
The zeros of 2x^3-4x^2-32x+64=0 are x=2,x=-4,x=4